Spatial transport of a neutral doublet

Summary: The two neutral polarizations of a balanced five-dimensional rotation carry a nonzero traceless geometric curvature when its physical null axis varies, with an exact spatial kinetic-energy decomposition.

A neutral mode can respond to changing spatial geometry

Rotation-neutral spatial modes identifies a two-component neutral subspace for the sum of two equal spatial rotation planes. This note derives how those modes respond when the rotation's physical null axis changes. Their geometric response is nonzero even though the balanced tangential rotation annihilates them.

The axis is an ordinary spatial direction in a five-dimensional tangent space of the one infinite manifold. Its orientation is described by a unit vector $n\in S^4$. This sphere is a set of possible directions, not an additional physical space. When $n=n(x)$, its variation occurs over actual spatial positions $x$; all derivatives below are pulled back to those positions. The two complex amplitudes describe polarization of the same field $\Psi_\infty$, not two extra spatial dimensions or mediator particles.

The explicit kinetic calculations use a flat Euclidean spatial chart and ordinary derivatives. A curved sector requires its metric and spin connection in the same projection.

This is a conditional projected-kinetic mechanism. It does not identify the resulting connection with the observed weak interaction or its trace with electric charge. The physical mode selection and electromagnetic current remain to be derived.

The neutral projector is fixed by the physical null axis

Use the five Hermitian Clifford matrices

$$\gamma=(\sigma_1\otimes I,\sigma_2\otimes I, \sigma_3\otimes\sigma_1,\sigma_3\otimes\sigma_2, \sigma_3\otimes\sigma_3),\qquad \{\gamma_a,\gamma_b\}=2\delta_{ab}I_4. \tag{1}$$

For a pure normalized source polarization $z$, put

$$n_a=z^\dagger\gamma_a z,\quad J_{ab}=z^\dagger(-i\gamma_a\gamma_b)z\ (a<b),\quad S=\sum_{a<b}J_{ab}(-i\gamma_a\gamma_b). \tag{2}$$

In an adapted spatial frame, $S=H_{12}+H_{34}$ and $n\cdot\gamma=H_{12}H_{34}$. Spatial covariance therefore gives the neutral projector in every frame:

$$\boxed{P(n)=\frac{I-n\cdot\gamma}{2} =I-\frac{S^2}{4},\qquad SP=0,\qquad \operatorname{rank}P=2.} \tag{3}$$

Thus $P$ depends only on the null axis. Changing the two-plane complex structure at fixed oriented axis can change $S$ while keeping the same neutral subspace. The null axis is relational geometric data, not an inherent partition of the particle's five ordinary dimensions.

Choose a local $4\times2$ matrix $W(n)$ with

$$W^\dagger W=I_2,\qquad WW^\dagger=P, \qquad \Phi(x)=W(n(x))b(x). \tag{4}$$

For an explicit frame near $n=e_5$, let $E$ contain orthonormal negative $\gamma_5$ eigenvectors. Then

$$W(n)=\sqrt{\frac{2}{1+n_5}}\,P(n)E,\qquad n_5\ne-1. \tag{5}$$

Indeed $E^\dagger P E=(1+n_5)I_2/2$. Another patch is needed at the opposite axis; a failure of this coordinate formula is not a singularity of the physical projector.

Noncommuting transport follows directly from projection

Define the anti-Hermitian local connection and its curvature by

$$\mathcal A=W^\dagger dW,\qquad \mathcal F=d\mathcal A+\mathcal A\wedge\mathcal A =W^\dagger(dP\wedge dP)W. \tag{6}$$

The last identity follows by differentiating $W^\dagger W=I$ and inserting $I=P+(I-P)$ into $dW^\dagger\wedge dW$. In coordinates,

$$\mathcal F_{ij}=W^\dagger[\partial_iP,\partial_jP]W. \tag{7}$$

At the north axis, use $n=(y_1,y_2,y_3,y_4,\sqrt{1-|y|^2})$. There $\partial_iP=-\gamma_i/2$, so for distinct $i,j\leq4$,

$$\boxed{\mathcal F_{ij}=\frac12W^\dagger\gamma_i\gamma_jW, \qquad\operatorname{tr}\mathcal F_{ij}=0, \qquad\|\mathcal F_{ij}\|_F^2=\frac12.} \tag{8}$$

The projected bivectors span three independent anti-Hermitian matrices. For example, Clifford multiplication gives

$$[\mathcal F_{12},\mathcal F_{13}]=-\mathcal F_{23}\ne0. \tag{9}$$

Consequently spatial changes in different directions can act in different, noncommuting ways on the two neutral amplitudes. This is an $\mathfrak{su}(2)$ curvature. Its vanishing trace holds globally: in $\operatorname{tr}P[\gamma_a,\gamma_b]$, both the two-gamma commutator trace and the relevant three-gamma trace vanish.

A change of local polarization basis $W\mapsto WU$, $b\mapsto U^\dagger b$ with $U\in U(2)$ leaves the physical field unchanged and gives

$$\mathcal A\mapsto U^\dagger\mathcal A U+U^\dagger dU, \qquad \mathcal F\mapsto U^\dagger\mathcal F U. \tag{10}$$

A full four-component rotating frame may have zero curvature. Projection onto the two-component subspace retains the geometry of its motion relative to its complement, producing (6). It becomes a physical reduced response only when a justified constraint or separated mode band makes that projection appropriate. A coordinate change alone does not create a force.

The exact spatial kinetic term

For a field satisfying (4), decompose its derivative into orthogonal pieces:

$$\partial_i\Phi=W D_i b+(I-P)(\partial_iW)b, \qquad D_i b=\partial_i b+\mathcal A_i b. \tag{11}$$

Their squared norms add. Since $\partial_iP=-\gamma(\partial_i n)/2$ and the Clifford square is $\gamma(v)^2=|v|^2I$, one obtains

$$\boxed{ \int\|\nabla\Phi\|^2dx =\int\left[\sum_i\|D_i b\|^2 +\frac14\sum_i|\partial_i n|^2\,b^\dagger b\right]dx.} \tag{12}$$

The coefficient of the extra positive term is fixed by the same kinetic energy. It is not a second fitted coupling. Both the orientation field $n(x)$ and the amplitude $b(x)$ must vary in the reciprocal field equations when both are dynamical. Freezing $n$ is a specified-background calculation. The source's remaining energy and kinetic terms must also be retained.

For a slow time-dependent constrained mode, the Berry term similarly gives $i b^\dagger\dot b+i b^\dagger\mathcal A_t b$. The geometrical part of its amplitude transport is $\dot b=-\mathcal A_t b$. Around a small oriented loop of area $\delta S$ in the $i,j$ directions,

$$b_{\rm final}=\left(I-\mathcal F_{ij}\delta S +o(\delta S)\right)b_{\rm initial}. \tag{13}$$

This is a concrete response of the neutral modes to moving geometry. A constant axis has no such curvature on physical space. If the actual orientation texture varies through only one parameter, the pulled-back two-form also vanishes locally. Nonzero geometric curvature requires variation in at least two independent orientation directions.

Relation to the first-order spatial Dirac operator

Equation (12) starts from a positive spatial gradient energy. The master Dirac kinetic term is first order, so its projection requires a separate calculation. For every ordinary spatial vector $v$, the projector (3) gives

$$\boxed{W^\dagger\gamma(v)W=-(n\cdot v)I_2.} \tag{12a}$$

To see this, decompose $v=(n\cdot v)n+v_\perp$. The first Clifford factor acts as $-I$ on the selected eigenspace. The second anticommutes with $n\cdot\gamma$ and therefore maps it into its complement. For $D=-i\gamma_a\partial_a$ this yields

$$W^\dagger D(Wb)=i n_a\partial_a b -iW^\dagger\gamma_a(\partial_aW)b. \tag{12b}$$

The leading momentum symbol is $-(n\cdot p)I_2$. It is not a closed five-dimensional Clifford symbol. The actual field can still extend in all five spatial directions; the point is that its omitted components carry essential first-order responses in the perpendicular directions.

This difference is already exact for a constant axis and constant $W$:

$$\boxed{ W^\dagger D^2W-(W^\dagger D W)^2 =W^\dagger D(I-P)DW =-\big[\Delta_x-(n\cdot\nabla)^2\big]I_2.} \tag{12c}$$

At a real momentum its eigenvalue is $|p|^2-(n\cdot p)^2\geq0$. Thus projecting the squared Dirac kinetic operator retains contributions that squaring the projected first-order operator would discard. This supplies an explicit complement term to calculate in a physical reduction. Equation (12) is a valid gradient-energy identity, but applying it as the particle's effective dynamics requires deriving that reduction, its gap assumptions and the retained response of the same field.

Rotation-neutral Dirac fields in physical space provides a complementary full-field construction. Its position-dependent lower component retains the omitted spin weights while orbital motion keeps the total spatial rotation exactly neutral.

A controlled finite-gap spatial motion

A solvable full four-component evolution tests the projection against its omitted modes. Let $S_0=H_{12}+H_{34}$, $G=H_{15}/2$, and prescribe a uniform physical rotation of the source,

$$R(t)=e^{-i\nu tG},\qquad H(t)=\Delta R(t)S_0R(t)^\dagger,\qquad \Delta>0. \tag{13a}$$

The eigenvalues are $(-2\Delta,0,0,2\Delta)$; this calculation assumes an initially occupied neutral band. The prescribed source motion is a background benchmark, not a completed reciprocal equation for that source. The neutral subspace moves as $R(t)P_0R(t)^\dagger$. In its rotating frame,

$$i\dot\chi=(\Delta S_0-\nu G)\chi, \qquad \Phi(t)=R(t)e^{-it(\Delta S_0-\nu G)}\Phi(0). \tag{13b}$$

Here $P_0GP_0=0$, but $G$ couples the doublet to the two charged branches. Each coupled pair has off-diagonal magnitude $\nu/2$ and energy difference $2\Delta$. The exact probability of leaving the moving neutral subspace is consequently, for any normalized initial neutral polarization,

$$\boxed{p_{\rm out}(t)= \frac{\nu^2}{4\Delta^2+\nu^2} \sin^2\!\left(t\sqrt{\Delta^2+\frac{\nu^2}{4}}\right) \leq\frac{\nu^2}{4\Delta^2+\nu^2}.} \tag{13c}$$

This derives a controlled regime $|\nu|\ll\Delta$ for the projected transport. A gap does not make omitted modes identically absent. At a full physical circle $T=2\pi/|\nu|$, the moving spin frame returns as $R(T)=-I$, and the slow-limit neutral transport has holonomy $-I$. Virtual transitions also shift the accumulated neutral phases at finite speed; a small leakage probability alone does not bound that phase error. The circle illustrates that a curve can have nontrivial holonomy even though the pullback of a curvature two-form to a single curve vanishes.

A global check on the neutral bundle

For the convention (1), $\gamma_1\gamma_2\gamma_3\gamma_4\gamma_5=-I$. Expanding (3) gives the coordinate-free trace identity

$$\operatorname{tr}(\mathcal F\wedge\mathcal F) =\operatorname{tr}\big(P(dP)^4\big) =\frac18\epsilon_{eabcd}n_e\, dn_a\wedge dn_b\wedge dn_c\wedge dn_d =3\,\mathrm{vol}_{S^4}. \tag{14}$$

Here the outward orientation is used, so $\mathrm{vol}_{S^4}=\epsilon_{eabcd}n_e\,dn_a\wedge dn_b\wedge dn_c\wedge dn_d/4!$. Since $\operatorname{vol}(S^4)=8\pi^2/3$, the anti-Hermitian Chern–Weil convention $c_2=[\operatorname{tr}(\mathcal F\wedge\mathcal F) -\operatorname{tr}\mathcal F\wedge\operatorname{tr}\mathcal F]/(8\pi^2)$ yields

$$\int_{S^4}c_2=1,\qquad \operatorname{tr}\mathcal F=0. \tag{15}$$

Reversing orientation or taking the complementary doublet changes the second Chern number's sign. Its nonzero magnitude confirms that the curvature cannot be removed by choosing another polarization basis. It does not require a physical orientation texture to cover the whole sphere.

This Clifford-projector bundle is the mathematical structure commonly called a Yang monopole. An independent experimental realization in a controlled parameter space is described by Sugawa and colleagues. That realization supports the mathematical comparison; it is not evidence that the IDWT particle assignment or spatial coupling proposed here occurs in nature. The present interpretation uses orientations of an actual spatial axis, with the derivation supplied explicitly above.

Dressed spatial source vertices derives the complete source and its normalization after eliminating those omitted components, including exact finite-frequency continuity identities.

What this supplies to the coupling problem

The same higher geometric source can now support two precise statements: its balanced tangential rotation annihilates the doublet, while spatial changes of its null axis induce nonzero, noncommuting transport within it. This provides an explicit example of neutrality in one geometric operation coexisting with another response, without inserting an arbitrary second matrix to manufacture that distinction.

There are still essential physical selection questions. The zero doublet is the middle eigenspace of $S$, not its unique ground state. The scalar self-binding model admits neutral and nonneutral polarizations at the same energy. A physical constraint, isolated populated band or dynamical selection must therefore be derived before (12) closes the actual field problem. At finite gaps, omitted modes contribute corrections and possible transitions. The moving-mode equation retains their exact terms.

Furthermore, the full six-dimensional source contains a mixed contribution acting nontrivially on this doublet. Its absence from a photon response cannot be assumed. The descending source projection must identify the physical current. Zero determinant curvature in (8) is not a proof of zero electric charge, and nonzero $\mathfrak{su}(2)$ curvature is not a calculation of weak chirality, coupling strength or scattering rate. Those identifications require the same field action and the occupied spatial particle solutions.

Verification

spatial_coupling_mechanism_checks.py checks (6) using differentiated explicit frames and projector commutators, local basis covariance, noncommuting curvature components, the exact full kinetic energy (12), and independent spatial finite differences. Consecutive projector overlap transport around shrinking loops verifies (13) without using the curvature formula in the evolution. Full rotating-source evolution checks the finite-gap leakage formula against the original time-dependent field equation. Independent momentum symbols verify (12a)–(12c), including the full omitted-mode contribution. An antisymmetrized Clifford trace evaluates all 24 terms of (14) in independently rotated tangent frames.

Related

• Rotation-neutral spatial modes — supplies the neutral doublet and its nonzero mixed-source response.
• Rotations of nested spatial sectors — distinguishes compatible motion of a nested chain from freezing each member.